Question

Решите плиз номера под 4,5,6 номерами кроме последнего варианта

Answer 1

$( sqrt{7} - sqrt{3})^2=7-2 sqrt{21}+3=10-2 sqrt{21} \ (5+ sqrt{3})^2 =25+10 sqrt{3}+3=28+10 sqrt{3} \ (5- sqrt{3})^2 =25-10 sqrt{3}+3=28-10 sqrt{3} \ ( sqrt{5}+sqrt{3})^2 =5+2sqrt{15}+3=8+2sqrt{15} \ ( sqrt{5}-sqrt{3})^2 =5-2sqrt{15}+3=8-2sqrt{15} \ 0,5 sqrt{0,01}- frac{1}{26} *sqrt{169}= frac{1}{2}*frac{1}{10}-frac{1}{2} =frac{1}{20}-frac{1}{2} =-frac{9}{20} \ 7 sqrt{0,04}+frac{1}{6} sqrt{144}=7frac{1}{5}+2=frac{7}{5}+2=3,4$
$7 sqrt{0,04}-frac{1}{6} sqrt{144}=7 sqrt{ frac{1}{25} } -frac{1}{6}*12=frac{7}{5}-2=-0,6 \ 0,5 sqrt{0,04} +frac{1}{16} sqrt{100} =frac{1}{2}*frac{1}{5}+frac{1}{8}*5=frac{29}{40}=0,725 \ 5 sqrt{0,04}- frac{1}{16} sqrt{100}=5*frac{1}{5}-frac{1}{8}*5=frac{3}{8}=0,375 \ 2sqrt{2}+2sqrt{50}-2sqrt{32}=2sqrt{2}+10sqrt{2}-8sqrt{2}=4sqrt{2} \ 7sqrt{2}-2sqrt{50}+2sqrt{32}=7sqrt{2}-10sqrt{2}+8sqrt{2}=5sqrt{2}$
$12sqrt{2}+6sqrt{50}-3sqrt{32}= 12sqrt{2}+30sqrt{2}-12sqrt{2}=30sqrt{2} \ 10sqrt{2}-3sqrt{50}-2sqrt{32}=10sqrt{2}-15sqrt{2} -8sqrt{2}=-13sqrt{2} \ 8sqrt{2}-3sqrt{50}+2sqrt{32}=8sqrt{2}-15sqrt{2}+8sqrt{2}=sqrt{2}$